| Are A and B both on the same side of plane M? If so, then
the following may give the answer. Let B' be the "mirror
image" of B through M, i.e., drop a perpendicular from point
B to plane M and extend it its own length to the other side
of M and call that point B'. So if the perpendicular from B
to M intersects plane M at point Q, then BQB' is a straight
line and BQ = QB'. Now connect A and B' with a staright
line. Will the intersection of AB' with plane M be the
desired point?
Dan
/M
B /
\ /
\ /
\ /
A------------/------B'
/ P?
/
|
| Another pair for you to ponder:
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The focus F and the directrix d1d2 of a parabola are given, but not the
curve itself. Also given is a straight line m that would intersect the
parabola if it were present. Using straightedge and compass, determine
a point P on the line m where the parabola would cross it.
-----------------------------------------------------------------------
If no vertex of a simple polyhedron is incident with exactly 3 edges,
prove that at least 8 of the faces must be triangles.
Thanks in advance,
Steve Barkley.
|
| Euler says: G = E + 2, where E is # edges
and G is # faces & vertices
I say: G = T + X, where T is # triangles & vertices of deg 3.
and X are the rest.
4*E >= 3*T + 4*X = 4*G - T = 4*E + 8 - T
i.e. T >= 8.
Since there are no vertices of degree 3, there must be at least 8
triangles.
|
| Note 1005.2 gets one point of the locus (are there more?).
Note 1005.1 is wrong. Consider M as the xy plane, and the points
A = (-1,0,1), B = (2,0,2):
z
^ B
| /|
A | / | C = (-1,0,0)
|\|/ | D = (2,0,0)
---C-O---D-> x
The triangles ACO and BDO are similar, so (OC)/(AC) = (OD)/(BD).
Note 1005.1 implies that for points P = (0,y,0), ACP and BDP are
similar, so that (PC)/(AC) = (PD)/(BD). But
(OP)^2 + (OC)^2 (PC)^2 (PD)^2 (OP)^2 + (OD)^2
--------------- = ------ = ------ = ---------------
(AC)^2 (AC)^2 (BD)^2 (BD)^2
Now subtracting [(OC)/(AC)]^2 = [(OD)/(BD)]^2 gives:
(OP)^2 (OP)^2
------ = ------
(AC)^2 (BD)^2
So that (AC) = (BD), and the perpendiculars from A and B to the plane
must be the same length. Hmmm...
- - - - - - - - -
The points P satisfy: (PC)/(AC) = (PD)/(BD), or (PC) = R*(PD), where
R = (AC)/(BD). Thus, the original problem can be reduced to the following:
Given two points C and D in the plane, and a positive real R, find
the locus of points P in the plane such that (PC) = R*(PD).
Without loss of generality, we can assume C = (R,0), and D = (1,0)
(by rotating and scaling the plane). Now, let P = (x,y) so that:
sqrt((x-R)^2 + y^2) = R sqrt((x-1)^2 + y^2)
...
2 2 R x 2
y = ----- - x
R + 1
Which is some moderately strange curve.
|